Tchebyshev inequality for Measure theory.












0












$begingroup$


Suppose $f$ integrable on measurable set $E$ and $f(x)$ $geq$ $0$ on $E$,



And for any $a in mathbb{R}^+$, we define:



$E_a$ = ${x in E$ $vert$ $f(x) > a}$. (which is measurable by basic principles)



Show that $m(E_a) leq frac{1}{a} int_E f(x)$. ($m$ is Leb Meas)



here is what I have:



Pf:



Since



$a$ $chi_{E_a}$ $leq f(x)$ ($because$ of the fact: $x in E_a$)



by integrating both sides over $E$ we obtain



$a int_E chi_{E_a}$ $leq$ $int_E f(x)$



and by definition, we rewrite the LHS and have:



$m(E_a) leq frac{1}{a} int_E f(x)$



does this work?



a classmate said I have to consider an infinite case too? but then



$int_{E_a} a chi_{E_a cap [-n,n]}$ goes off to infinity?



not sure I get this part, is my part correct? thanks in advance!






real-analysis measure-theory lebesgue-integral lebesgue-measure







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    Suppose $f$ integrable on measurable set $E$ and $f(x)$ $geq$ $0$ on $E$,



    And for any $a in mathbb{R}^+$, we define:



    $E_a$ = ${x in E$ $vert$ $f(x) > a}$. (which is measurable by basic principles)



    Show that $m(E_a) leq frac{1}{a} int_E f(x)$. ($m$ is Leb Meas)



    here is what I have:



    Pf:



    Since



    $a$ $chi_{E_a}$ $leq f(x)$ ($because$ of the fact: $x in E_a$)



    by integrating both sides over $E$ we obtain



    $a int_E chi_{E_a}$ $leq$ $int_E f(x)$



    and by definition, we rewrite the LHS and have:



    $m(E_a) leq frac{1}{a} int_E f(x)$



    does this work?



    a classmate said I have to consider an infinite case too? but then



    $int_{E_a} a chi_{E_a cap [-n,n]}$ goes off to infinity?



    not sure I get this part, is my part correct? thanks in advance!






    real-analysis measure-theory lebesgue-integral lebesgue-measure







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Suppose $f$ integrable on measurable set $E$ and $f(x)$ $geq$ $0$ on $E$,



      And for any $a in mathbb{R}^+$, we define:



      $E_a$ = ${x in E$ $vert$ $f(x) > a}$. (which is measurable by basic principles)



      Show that $m(E_a) leq frac{1}{a} int_E f(x)$. ($m$ is Leb Meas)



      here is what I have:



      Pf:



      Since



      $a$ $chi_{E_a}$ $leq f(x)$ ($because$ of the fact: $x in E_a$)



      by integrating both sides over $E$ we obtain



      $a int_E chi_{E_a}$ $leq$ $int_E f(x)$



      and by definition, we rewrite the LHS and have:



      $m(E_a) leq frac{1}{a} int_E f(x)$



      does this work?



      a classmate said I have to consider an infinite case too? but then



      $int_{E_a} a chi_{E_a cap [-n,n]}$ goes off to infinity?



      not sure I get this part, is my part correct? thanks in advance!






      real-analysis measure-theory lebesgue-integral lebesgue-measure







      share|cite|improve this question











      $endgroup$




      Suppose $f$ integrable on measurable set $E$ and $f(x)$ $geq$ $0$ on $E$,



      And for any $a in mathbb{R}^+$, we define:



      $E_a$ = ${x in E$ $vert$ $f(x) > a}$. (which is measurable by basic principles)



      Show that $m(E_a) leq frac{1}{a} int_E f(x)$. ($m$ is Leb Meas)



      here is what I have:



      Pf:



      Since



      $a$ $chi_{E_a}$ $leq f(x)$ ($because$ of the fact: $x in E_a$)



      by integrating both sides over $E$ we obtain



      $a int_E chi_{E_a}$ $leq$ $int_E f(x)$



      and by definition, we rewrite the LHS and have:



      $m(E_a) leq frac{1}{a} int_E f(x)$



      does this work?



      a classmate said I have to consider an infinite case too? but then



      $int_{E_a} a chi_{E_a cap [-n,n]}$ goes off to infinity?



      not sure I get this part, is my part correct? thanks in advance!






      real-analysis measure-theory lebesgue-integral lebesgue-measure




      real-analysis measure-theory lebesgue-integral lebesgue-measure






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 10 '18 at 23:23







      Hossien Sahebjame

















      asked Dec 10 '18 at 22:34









      Hossien SahebjameHossien Sahebjame

      888




      888






















          1 Answer
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          $begingroup$

          everything looks fine, you do not have to consider the infinite case since if the integral of f is not finite, then you have the claim trivially. If the integral of the indicator of function over the set you are given is infinte, then by monotonicity of the integral also the integral of f is infinite and again it holds :D






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            awesome thanks so much!! I see so the infinite case is the trivial case of the inequality? so for my finite case I had, that looks correct? I just need to include the infinite and say why it trivially holds, thanks@
            $endgroup$
            – Hossien Sahebjame
            Dec 10 '18 at 23:11










          • $begingroup$
            but I see what you mean, so this proof would suffice for the measure theoretic Tchebyshev inequality?
            $endgroup$
            – Hossien Sahebjame
            Dec 10 '18 at 23:12










          • $begingroup$
            I think so, anyway look at Markov inequality which is a general version of the tchebichev :D
            $endgroup$
            – anonymous
            Dec 12 '18 at 10:22










          • $begingroup$
            awesome thanks!
            $endgroup$
            – Hossien Sahebjame
            Dec 12 '18 at 15:59











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          $begingroup$

          everything looks fine, you do not have to consider the infinite case since if the integral of f is not finite, then you have the claim trivially. If the integral of the indicator of function over the set you are given is infinte, then by monotonicity of the integral also the integral of f is infinite and again it holds :D






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            awesome thanks so much!! I see so the infinite case is the trivial case of the inequality? so for my finite case I had, that looks correct? I just need to include the infinite and say why it trivially holds, thanks@
            $endgroup$
            – Hossien Sahebjame
            Dec 10 '18 at 23:11










          • $begingroup$
            but I see what you mean, so this proof would suffice for the measure theoretic Tchebyshev inequality?
            $endgroup$
            – Hossien Sahebjame
            Dec 10 '18 at 23:12










          • $begingroup$
            I think so, anyway look at Markov inequality which is a general version of the tchebichev :D
            $endgroup$
            – anonymous
            Dec 12 '18 at 10:22










          • $begingroup$
            awesome thanks!
            $endgroup$
            – Hossien Sahebjame
            Dec 12 '18 at 15:59
















          1












          $begingroup$

          everything looks fine, you do not have to consider the infinite case since if the integral of f is not finite, then you have the claim trivially. If the integral of the indicator of function over the set you are given is infinte, then by monotonicity of the integral also the integral of f is infinite and again it holds :D






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            awesome thanks so much!! I see so the infinite case is the trivial case of the inequality? so for my finite case I had, that looks correct? I just need to include the infinite and say why it trivially holds, thanks@
            $endgroup$
            – Hossien Sahebjame
            Dec 10 '18 at 23:11










          • $begingroup$
            but I see what you mean, so this proof would suffice for the measure theoretic Tchebyshev inequality?
            $endgroup$
            – Hossien Sahebjame
            Dec 10 '18 at 23:12










          • $begingroup$
            I think so, anyway look at Markov inequality which is a general version of the tchebichev :D
            $endgroup$
            – anonymous
            Dec 12 '18 at 10:22










          • $begingroup$
            awesome thanks!
            $endgroup$
            – Hossien Sahebjame
            Dec 12 '18 at 15:59














          1












          1








          1





          $begingroup$

          everything looks fine, you do not have to consider the infinite case since if the integral of f is not finite, then you have the claim trivially. If the integral of the indicator of function over the set you are given is infinte, then by monotonicity of the integral also the integral of f is infinite and again it holds :D






          share|cite|improve this answer









          $endgroup$



          everything looks fine, you do not have to consider the infinite case since if the integral of f is not finite, then you have the claim trivially. If the integral of the indicator of function over the set you are given is infinte, then by monotonicity of the integral also the integral of f is infinite and again it holds :D







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 10 '18 at 22:57









          anonymousanonymous

          361




          361












          • $begingroup$
            awesome thanks so much!! I see so the infinite case is the trivial case of the inequality? so for my finite case I had, that looks correct? I just need to include the infinite and say why it trivially holds, thanks@
            $endgroup$
            – Hossien Sahebjame
            Dec 10 '18 at 23:11










          • $begingroup$
            but I see what you mean, so this proof would suffice for the measure theoretic Tchebyshev inequality?
            $endgroup$
            – Hossien Sahebjame
            Dec 10 '18 at 23:12










          • $begingroup$
            I think so, anyway look at Markov inequality which is a general version of the tchebichev :D
            $endgroup$
            – anonymous
            Dec 12 '18 at 10:22










          • $begingroup$
            awesome thanks!
            $endgroup$
            – Hossien Sahebjame
            Dec 12 '18 at 15:59


















          • $begingroup$
            awesome thanks so much!! I see so the infinite case is the trivial case of the inequality? so for my finite case I had, that looks correct? I just need to include the infinite and say why it trivially holds, thanks@
            $endgroup$
            – Hossien Sahebjame
            Dec 10 '18 at 23:11










          • $begingroup$
            but I see what you mean, so this proof would suffice for the measure theoretic Tchebyshev inequality?
            $endgroup$
            – Hossien Sahebjame
            Dec 10 '18 at 23:12










          • $begingroup$
            I think so, anyway look at Markov inequality which is a general version of the tchebichev :D
            $endgroup$
            – anonymous
            Dec 12 '18 at 10:22










          • $begingroup$
            awesome thanks!
            $endgroup$
            – Hossien Sahebjame
            Dec 12 '18 at 15:59
















          $begingroup$
          awesome thanks so much!! I see so the infinite case is the trivial case of the inequality? so for my finite case I had, that looks correct? I just need to include the infinite and say why it trivially holds, thanks@
          $endgroup$
          – Hossien Sahebjame
          Dec 10 '18 at 23:11




          $begingroup$
          awesome thanks so much!! I see so the infinite case is the trivial case of the inequality? so for my finite case I had, that looks correct? I just need to include the infinite and say why it trivially holds, thanks@
          $endgroup$
          – Hossien Sahebjame
          Dec 10 '18 at 23:11












          $begingroup$
          but I see what you mean, so this proof would suffice for the measure theoretic Tchebyshev inequality?
          $endgroup$
          – Hossien Sahebjame
          Dec 10 '18 at 23:12




          $begingroup$
          but I see what you mean, so this proof would suffice for the measure theoretic Tchebyshev inequality?
          $endgroup$
          – Hossien Sahebjame
          Dec 10 '18 at 23:12












          $begingroup$
          I think so, anyway look at Markov inequality which is a general version of the tchebichev :D
          $endgroup$
          – anonymous
          Dec 12 '18 at 10:22




          $begingroup$
          I think so, anyway look at Markov inequality which is a general version of the tchebichev :D
          $endgroup$
          – anonymous
          Dec 12 '18 at 10:22












          $begingroup$
          awesome thanks!
          $endgroup$
          – Hossien Sahebjame
          Dec 12 '18 at 15:59




          $begingroup$
          awesome thanks!
          $endgroup$
          – Hossien Sahebjame
          Dec 12 '18 at 15:59


















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